m = 10

h_1(k) = (2^k mod (m + 1)) - 1
h_2(k) = (5^k mod (m + 1)) - 1
h_3(k) = (7^k mod (m + 1)) - 1

insert [2, 4]

i: 0123456789
   ---
W: 0000000000, insert(2)
W: 0011100000, insert(4)
W: 0011100010

strlst@ayaya ~ python3
Python 3.8.2 (default, Mar 24 2020, 03:08:36)
[GCC 9.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> def h_1(k):
...     return ((2**k) % 11) - 1
...
>>> def h_2(k):
...     return ((5**k) % 11) - 1
...
>>> def h_3(k):
...     return ((7**k) % 11) - 1
...
>>> numbers = [2, 4]
>>> [(h_1(n), h_2(n), h_3(n)) for n in numbers]
[(3, 2, 4), (4, 8, 2)]
>>> numbers2 = [3, 5, 6]
>>> [(h_1(n), h_2(n), h_3(n)) for n in numbers2]
[(7, 3, 1), (9, 0, 9), (8, 4, 3)]
>>>

